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1 year ago

What is the name for potassium oxide's structure? Give two properties of potassium oxide.

Chemistry

1 answer:

Anna71 [15]1 year ago

8 0

Answer:

- The name for the potassium oxide's structure is ionic.

Properties:

- High melting point.

- Soluble in water.

Explanation:

- The ionic structure it is formed by a cation (atom with positive charge) and an anion (atom with negative charge). In this case, potassium is the cation and the oxigen is the anion.

- Since potassium oxide is an ionic compound, it has a high melting point, because of the strong bonds. Also, it is soluble in polar solvents, like water, because its ions generate polarity in the molecule.

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How many molecules (not moles) of NH3 are produced from 5.25x10^-4 g of H2?

Oksi-84 [34.3K]

Answer:

not 100% but i think its 1.57x10^20

Explanation:

5.25x10^-4g / 2.016g

2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules

6 0

2 years ago

31. Adding an additional oxygen atom to 02 creates....

cricket20 [7]

It creates chlorofluorocarbons(CFCs) also would create biooxygen but in the multiple choice it only shows the CFCs

4 0

2 years ago

The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially

Crank

Answer: The rate constant is $0.334s^{-1}$

Explanation ;

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = age of sample = 4.26 min

a = initial amount of the reactant = 2.56 mg

a - x = amount left after decay process = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

$4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}$

$k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}$

$k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}$

Thus rate constant is [tex]0.334s^{-1}

4 0

2 years ago

3.604m + 104.29m + 3.1m + 17.41m. The sum expressed in the correct number of significant figures is

dedylja [7]

Answer:

128.4 m

Explanation:

3.604m + 104.29m + 3.1m + 17.41m

Add all the values

= 128.404 m

The significant figure rule for addition is for the sum to have the same number of decimal places as the value with the least number of decimal places. In the addition sentence 3.604m + 104.29m + 3.1m + 17.41m, the value with the least number of decimal places is 3.1, which has 1 decimal place. Therefore, we round our sum so that it also has 1 decimal place.

128.404 m

= 128.4 m

I hope this helps!

7 0

2 years ago

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com

FrozenT [24]

The mass percent of Pentane in solution is 16.49%

The mass percent of Hexane in solution is 83.51%

<u>Explanation</u>:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

From Raoult's law we know:

Pp = xp $\times$ Pp - vap = yp $\times$ Pt (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt (2)

Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt (3)

Substituting (1) into (3) we get:

(1-xp) $\times$ Ph - vap = (1 - yp) $\times$ xp $\times$ Pp - vap / yp (4)

Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)

Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%

Hexane mol% in solution: xh = 80.92%

Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol $\times$ 72.15 g/mol

= 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

= 69.737 g

Mass% of Pentane solution = 13.766/(13.766+69.737)

= 16.49%

Mass% of Hexane solution = 83.51%

8 0

3 years ago